获得达到所需数量所需的天数

Question

我目前有一个数据框，其中包含站点名称、降雨日期、降雨量（附示例）我有兴趣探索每个站点到达所需的天数（and/or 个月）达到一定的降水量。例如：

是否可以根据示例数据集获得上述输出？我最初的想法是单独过滤每个站，加入一个日历数据框，从该范围中提取最小值和最大值，计算它们之间的天数并使用 case_when 对它们进行分类。这种方法似乎有点令人费解，如有任何关于更好方法的指导，我们将不胜感激。

感谢您的建议！

示例数据集：

Example <- structure(list(Name.Station = c("Station A", "Station A", "Station A", 
                                        "Station A", "Station A", "Station B", "Station B", "Station B", 
                                        "Station C", "Station C", "Station C", "Station C"), Rainfall.Date = c("7/10/2020", 
                                                                                                               "8/12/2020", "8/01/2021", "25/06/2021", "26/10/2021", "7/01/2020", 
                                                                                                               "22/01/2020", "5/02/2020", "5/09/2020", "5/10/2020", "5/11/2020", 
                                                                                                               "5/12/2020"), Rainfall.Amount = c(210, 210, 208.47, 208.16, 203.67, 
                                                                                                                                                 227.49, 225, 222.54, 250, 250, 246.18, 245.15)), class = "data.frame", row.names = c(NA, 
                                                                                                                                                                                                                                      -12L))

Answer 1

by站你可以计算cumsum的降雨量大于mm的阈值。然后计算从开始日期到 cumsum 中最大日期的 seq 天数的 length。

不过，首先，您的日期格式应该正确。

Example <- transform(Example, Rainfall.Date=as.Date(Rainfall.Date, '%d/%m/%Y'))

do.call(rbind, by(Example, Example$Name.Station, \(x) {
  f <- \(mm, x.=x) {
    mx <- which.max(cumsum(x.$Rainfall.Amount) > mm)
    length(do.call(seq.Date, c(as.list(range(x.$Rainfall.Date[1:mx])), 1)))
  }
  ds <- seq.int(200, 1e3, 200)  ## sequence of 200, 400, ... , 1000mm
  r <- t(vapply(ds, f, 0))
  data.frame(Name.Station=el(x$Name.Station), `colnames<-`(r, paste0('d_', ds)))
}))
#           Name.Station d_200 d_400 d_600 d_800 d_1000
# Station A    Station A     1    63    94   262    385
# Station B    Station B     1    16    30     1      1
# Station C    Station C     1    31    62    92      1

注意： R >= 4.1 使用。

Answer 2

这是一个tidyverse方法：

library(dplyr)
library(tidyr)

Example %>%
  group_by(Name.Station) %>%
  mutate(Rainfall.Date = as.Date(Rainfall.Date, "%d/%m/%Y"),
         days = cumsum(c(1, diff(Rainfall.Date))),
         crainfall = cumsum(Rainfall.Amount),
         fi = (findInterval(crainfall, seq(0, 1000, 200)) -1) * 200) %>%
  pivot_wider(id_cols = Name.Station, names_from = fi, values_from = days, names_glue = {"days_to_{fi}_mm"}, values_fn = min)

# A tibble: 3 x 6
# Groups:   Name.Station [3]
  Name.Station days_to_200_mm days_to_400_mm days_to_600_mm days_to_800_mm days_to_1000_mm
  <chr>                 <dbl>          <dbl>          <dbl>          <dbl>           <dbl>
1 Station A                 1             63             94            262             385
2 Station B                 1             16             30             NA              NA
3 Station C                 1             31             62             92              NA

获得达到所需数量所需的天数

Obtaining the number of Days it takes to reach the desired amount

r

dplyr

lubridate